Let $C$ be the set of all Hahn Banach Extensions of a function. I have shown it to be non empty and closed. But how to show that $C$ is a convex set?
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Let $\mathcal A$ be a subspace of a Banach space $\mathcal B$, and assume $f,f':\mathcal B\to K$ are both extensions of $f_0:\mathcal A\to K$.
Then $\ g:=\lambda\cdot f\ +\ (1-\lambda)\cdot f'\ $ is also an extension of $f_0$.
If $\ \|f\|=\|f'\|=\|f_0\|\ $ and $\lambda\in [0,1]$, then for a vector $b\in\mathcal B$ with $\|b\|=1$, we have $$\matrix{|g(b)|&=&|\lambda\cdot f(b)+(1-\lambda)\cdot f'(b)|&\le \\ &\le&\lambda\cdot|f(b)|+(1-\lambda)\cdot|f'(b)| & \le \\ &\le& \lambda\cdot\|f\|+(1-\lambda)\cdot\|f'\| & = & \|f_0\|}$$ On the other hand, as $g|_{\mathcal A}=f_0$, we have $\|g\|\ge\|f_0\|$.
Berci
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1but how do you claim that $|g| = |\lambda\cdot f\ +\ (1-\lambda)\cdot f'|$ – User8976 Nov 25 '17 at 00:56
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I updated my answer with this. – Berci Nov 25 '17 at 10:26